The major product of the following reaction is:

Step 1: Identify the starting material

The starting material is 1,2-dihydronaphthalene (tetralin with one double bond in the ring) treated with $\ce{Br2}$ in the presence of ethanol ($\ce{EtOH}$).

Step 2: Bromonium ion formation

When $\ce{Br2}$ reacts with the alkene in the presence of an alcohol:

1. Bromonium ion ($\ce{Br^+}$) forms on the double bond → cyclic bromonium intermediate
2. The nucleophile (EtOH, not $\ce{Br^-}$) attacks the bromonium ion
3. EtOH attacks at the more electrophilic carbon (benzylic position)
4. Anti addition → trans product

Step 3: Regiochemistry and stereochemistry

• EtOH attacks at the benzylic carbon (C1 of the ring system, adjacent to aromatic ring) — more electrophilic due to benzylic stabilization
• $\ce{Br-}$ is incorporated at C2
• Anti addition → trans configuration of OEt and Br

Product = trans-1-ethoxy-2-bromotetralin (trans-2-bromo-1-ethoxytetralin)

Answer: Option (a)

Key Points to Remember:

• In ROH solvent, the alcohol acts as the nucleophile (instead of $\ce{Br-}$) → mixed product
• EtOH attacks the more electrophilic (benzylic) carbon
• Anti addition → trans relationship between OR and Br
• This reaction is used to make bromohydrin ethers